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Sav [38]
3 years ago
13

50.00 ml of nitric acid (of unknown

Chemistry
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

A)HNO_3 + NaOH → NaNO_3 + H2O

B) Explained below

C)volume of NaOH at the equivalence point will simply be the volume added to reach a pH value of 7.

D)the pH at the equivalence point is 7.

E)C(HNO3) = 0.1 × (volume of NaOH at equivalence point)/50

Explanation:

A) A symbolic equation for the reaction is;

HNO_3 + NaOH → NaNO_3 + H2O

B) HNO3 is a strong acid and NaOH is a strong base and as such they will both dissociate almost fully in a water solution to form ions. Thus, the reaction equation can now be rewritten in form of:

H(+) + OH(-) → H2O

From this reaction equation, it's clear that when sodium hydroxide (NaOH) was added, the positive hydrogen ions (H+) react with the negative hydroxyl ions (OH-) to form water (H2O).

Thus, the concentration [H+] will decrease because of both dilution and the decreasing amount of positive hydrogen ions (H+) in the solution. From definition, the pH is given by the formula;

pH = −log[H+]

Thus, pH increases as NaOH is added to HNO3 as given in the question.

C) The volume of NaOH at the equivalence point will simply be the volume added to reach a pH value of 7.

D) Since the given acid and base are very strong, it means that the solution at equivalence point is likely to be neutral. Thus, the pH at the equivalence point is 7.

E) Concentration of the nitric acid solution is given by;

C(HNO3) = (volume of NaOH at equivalence point × concentration of NaOH)/(volume of HNO3)

We are given;

concentration of NaOH) = 0.1 mol/L

Volume of HNO3 = 50 mL

Thus;

C(HNO3) = 0.1 × (volume of NaOH at equivalence point)/50

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